Patterns in Pascal's Triangle

Pascal's Triangle conceals a huge number of various patterns, many discovered by Pascal himself and even known before his time.

Pascal's Triangle is symmetric

In terms of the binomial coefficients, $C^{n}_{m} = C^{n}_{n-m}.$ This follows from the formula for the binomial coefficient

$\displaystyle C^{n}_{m}=\frac{n!}{m!(n-m)!}.$

It is also implied by the construction of the triangle, i.e., by the interpretation of the entries as the number of ways to get from the top to a given spot in the triangle.

Some authors even considered a symmetric notation (in analogy with trinomial coefficients)

$\displaystyle C^{n}_{m}={n \choose m\space\space s}$

where $s = n - m.$

The sum of entries in row $n$ equals $2^{n}$

This is Pascal's Corollary 8 and can be proved by induction. The main point in the argument is that each entry in row $n,$ say $C^{n}_{k}$ is added to two entries below: once to form $C^{n + 1}_{k}$ and once to form $C^{n + 1}_{k+1}$ which follows from Pascal's Identity:

"Pentatope" is a recent term. Regarding the fifth row, Pascal wrote that ... since there are no fixed names for them, they might be called triangulo-triangular numbers. Pentatope numbers exists in the $4D$ space and describe the number of vertices in a configuration of $3D$ tetrahedrons joined at the faces.

In the standard configuration, the numbers $C^{2n}_{n}$ belong to the axis of symmetry. Numbers $\frac{1}{n+1}C^{2n}_{n}$ are known as Catalan numbers.

Hockey Stick Pattern

In Pascal's words (and with a reference to his arrangement), In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding row from its column to the first, inclusive (Corollary 2). In modern terms,

Naturally, a similar identity holds after swapping the "rows" and "columns" in Pascal's arrangement: In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding column from its row to the first, inclusive (Corollary 3).

(2)

$C^{n + 1}_{m + 1} = C^{n}_{m} + C^{n - 1}_{m} + \ldots + C^{0}_{m},$

where the second index is fixed.

Parallelogram Pattern

(3)

$C^{n + 1}_{m} - 1 = \sum C^{k}_{j},$

where $k \lt n,$ $j \lt m.$ In Pascal's words: In every arithmetic triangle, each cell diminished by unity is equal to the sum of all those which are included between its perpendicular rank and its parallel rank, exclusively (Corollary 4). This is shown by repeatedly unfolding the first term in (1).

Fibonacci Numbers

If we arrange the triangle differently, it becomes easier to detect the Fibonacci sequence:

The successive Fibonacci numbers are the sums of the entries on sw-ne diagonals:

Sums of the Binomial Reciprocals

Squares

As I mentioned earlier, the sum of two consecutive triangualr numbers is a square: $(n - 1)n/2 + n(n + 1)/2 = n^{2}.$ Tony Foster brought up sightings of a whole family of identities that lead up to a square.